Discussion in 'The Projects Forum' started by whycanot, Aug 1, 2011.
Anyone can explain how is the circuit works?
Thanks in advanced
This is a comparator circuit using an op-amp (1/4 of the 741). There is a voltage divider consisting of the NTC thermistor and the variable resistor. When the temperature exceeds the setting determined by VR1, the output of U1 goes high which lights the LED and turns on transistor Q1, which operates the relay.
ETA: Although they don't tell you the value of the thermistor, it is most likely 10k ohms.
ETA: praondevou is correct. The output of U1 goes low to turn on the transistor and operate the relay.
Why do you open a new thread?
In the other thread ("Explanation needed") you inverted pin 2 and 3 of the 741 in your schematic. That's why I said it couldn't work.
Here it's a different story. The output of the 741 is normally HIGH. The transistor is not conducting, the relay not energized.
VR1 is adjusted so that pin 2 is normally lower in voltage than pin 3.
When the NTC heats up, it's resistance decreases. Therefore the voltage on pin 2 increases. When the voltage on pin 2 gets higher than on pin 3 the 741's output goes low, the transistor conducts and the relay is being energized.
What about the TSOP part of the diagram in your other thread.Is there an error too?
i open a new thread, cause my previous thread is not getting reply.
i hav 2 days more to understand these. sorry for making it messy.
im really happy to get u guys help. appreciate !
btw, what is the uses of R1 and R2 which connected to pin 3?
I was replying your other thread. I told you there was something wrong with the circuit.
R1 and R2 provide a fixed voltage of half the supply voltage to pin 3. The opamp works as a comparator. It compares the voltage levels on pin 3 and pin 2.
If the voltage on pin 2 is higher than on pin 3 the output voltage goes near zero volts. If the voltage on pin 2 is lower than on pin 3 the output voltage goes near the supply voltage (+).
Since the pnp transistor is conducting when you pull it's base towards zero, it will do so when the voltage on pin 2 is higher than on pin 3.
u means when the voltage on pin 2 is higher than on pin 3, the output which is pin 6 become near 0 volt, then the PNP transistor will be turned on?
so when pin 6 is 0 volt, why the LED can be turned ON ?
because you have current flowing from the emitter to base and collector.
base current is negative. please read this tutorial.